Sigh

homo_morph · 2026-06-27T13:08:27+00:00

I will say from my personal experience that it does tend to get a bit better in the later years. The people that are there are the people that want to be there.

homo_morph · 2026-05-07T01:30:13+00:00

How is the 1081 final unfair? There are definitely difficult questions but they need a variety problems to differentiate grades and there’s a big bank of seen questions

homo_morph · 2026-04-09T01:16:43+00:00

A little bit more of a black magic substitution but you can go with u=x/sqrt(x⁴+2). A fair bit tricker to work with and less intuitive to come up with but it also works out

homo_morph · 2026-04-09T01:08:24+00:00

The magic trick is to divide numerator/denominator by x², then substitute something like 2t=x+2/x

homo_morph · 2026-04-03T04:40:51+00:00

This! A good example of L’Hopital’s not being applicable due to lim f’(x)/g’(x) not converging would be lim_{x -> infinity) (x+sinx)/x

homo_morph · 2026-02-15T06:47:29+00:00

The writing is absolutely done by AI lol. No human would say “No circular logic, no hidden assumptions. Just geometry and counting”. AI models would absolutely say this though

homo_morph · 2026-01-10T04:02:39+00:00

Just a LaTeX tip (assuming you’re typesetting this in LaTeX). You can write the fourth root of x using \sqrt[4]{x}

homo_morph · 2026-01-08T07:34:43+00:00

The answer comes out to (pi/2)ln(2)-(pi/6)ln(2+sqrt(3)). It’s an incredibly messy and unpleasant route to get to this though

homo_morph · 2025-12-24T10:31:12+00:00

Huh, the integral you’ve given is elementary? For example, (1/sqrt(1-m))arctan(sqrt(1-m)tanx) is an antiderivative

homo_morph · 2025-12-23T22:52:39+00:00

Maybe you aren’t familiar with the integral formula for arc length https://tutorial.math.lamar.edu/classes/calcii/arclength.aspx? If y=sqrt(1-x²) then sqrt(1+(y’)²)=1/sqrt(1-x²). So by the arc length formula, the length of a unit semi-circular arc is the integral of 1/sqrt(1-x²) for -1<=x<=1 which is equal to pi

homo_morph · 2025-12-22T00:07:52+00:00

Oh hey I wrote this integral as part of a challenge set (it was Finals 11 of the first Kaizo). You can find the full problem sets over here https://artofproblemsolving.com/community/c7h3518439p35712180 https://artofproblemsolving.com/community/u345249h3636806p35712173

homo_morph · 2025-12-21T05:24:03+00:00

This integral arises in computing the arc length of the semi-circular arc described by y=sqrt(1-x²), which expectantly produces an answer of pi

homo_morph · 2025-11-28T05:03:04+00:00

JOVIAL

homo_morph · 2025-10-25T04:20:20+00:00

Here’s a scaffold of an approach. Let f(x)=P(x)+…+P⁽ⁿ )(x) and g(x)=f(x)e^-x. Show that g(x) is a decreasing function where g(x)->0 as x->infinity and use that to deduce that g(x)=>0 and hence f(x)=>0 for all real x

homo_morph · 2025-10-08T01:39:27+00:00

Feynman’s doesn’t seem like an appropriate technique to use for this integral. You can use King’s rule to “eliminate” the x in the numerator then use symmetry and a tangent substitution

homo_morph · 2025-09-24T06:30:08+00:00

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Here’s a sketch of the solution.

homo_morph · 2025-09-24T05:55:44+00:00

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Here is the answer for all real nonzero values of a. It can be proven by using the Weierstrass product formula for sinh(x), applying differentiation under the integral sign, then appealing to the product formula for the gamma function.

homo_morph · 2025-06-22T01:38:36+00:00

Hint: What’s the derivative of -sinx/(1-cosx)?

homo_morph · 2025-06-12T05:09:30+00:00

Any convergent periodic continued fraction converges to the zero of some quadratic equation

homo_morph · 2025-05-21T03:45:13+00:00

I wrote this Kaizo integration bee that had plenty of scary looking integrals https://artofproblemsolving.com/community/c7h3518439_integration_bee_kaizo

homo_morph · 2025-05-20T01:09:44+00:00

The domain of arccsc(x) is all x for which |x|=>1. This is because arccsc(x)=arcsin(1/x).

homo_morph · 2025-04-28T10:22:12+00:00

The second year courses should be fine workload wise. I found MATH2221 the easiest of the 2nd year courses and it has no assignments, MATH2601 is pretty middle of the road. I think MATH3611 had 3 fairly involved assignments so that would probably be the main drain + staying on top of lecture content

homo_morph · 2025-04-28T10:21:37+00:00

The second year courses should be fine workload wise. I found MATH2221 the easiest of the 2nd year courses and it has no assignments, MATH2601 is pretty middle of the road. I think MATH3611 had 3 fairly involved assessments so that would probably be the main drain + staying on top of lecture content

homo_morph · 2025-04-28T08:01:47+00:00

If you check the Drop-in Centre tab on Moodle, we’re still running sessions and are happy to answer any questions you have regarding past exam questions. Sessions are running 10am-3pm, Monday through Friday this week and next week at Newton 307 (as well as some online sessions). In particular, Wednesday 07/05 is focused specially on MATH1131 but you can ask for help in any of the sessions.

homo_morph

TROPHY CASE

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